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Differentiate Each of the Following from First Principle: E √ a X + B - Mathematics

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प्रश्न

Differentiate each of the following from first principle:

\[e^\sqrt{ax + b}\]

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उत्तर

\[\left( x \right) \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[\frac{d}{dx}\left( e^\sqrt{ax + b} \right) = \lim_{h \to 0} \frac{e^\sqrt{ax + ah + b} - e^\sqrt{ax + b}}{h}\]
\[ = a \lim_{h \to 0} \frac{e^\sqrt{ax + ah + b} - e^\sqrt{ax + b}}{\left( ax + ah + b \right) - \left( ax + b \right)}\]
\[ = a \lim_{h \to 0} \frac{e^\sqrt{ax + b} \left( e^\sqrt{ax + ah + b} - \sqrt{ax + b} - 1 \right)}{\left( \sqrt{\left( ax + ah + b \right)} \right)^2 - \left( \sqrt{\left( ax + b \right)} \right)^2}\]
\[ = a e^\sqrt{ax + b} \lim_{h \to 0} \frac{e^\sqrt{ax + ah + b} - \sqrt{ax + b} - 1}{\left( \sqrt{\left( ax + ah + b \right)} - \sqrt{\left( ax + b \right)} \right)\left( \sqrt{\left( ax + ah + b \right)} + \sqrt{\left( ax + b \right)} \right)}\]
\[ = a e^\sqrt{ax + b} \lim_{h \to 0} \frac{e^\sqrt{ax + ah + b} - \sqrt{ax + b} - 1}{\sqrt{\left( ax + ah + b \right)} - \sqrt{\left( ax + b \right)}} \lim_{h \to 0} \frac{1}{\sqrt{\left( ax + ah + b \right)} + \sqrt{\left( ax + b \right)}}\]
\[ = a e^{{}^\sqrt{ax + b}} \left( 1 \right)\frac{1}{2\sqrt{ax + b}}\]
\[ = \frac{a e^{{}^\sqrt{ax + b}}}{2\sqrt{ax + b}}\]

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अध्याय 30: Derivatives - Exercise 30.2 [पृष्ठ २६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.2 | Q 3.1 | पृष्ठ २६

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